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            <h1 id="seo-header">『算法-ACM竞赛-图论』二分图-二分图的定义及其判断定</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-图论』二分图-二分图的定义及其判断定"><a href="#『算法-ACM-竞赛-图论』二分图-二分图的定义及其判断定" class="headerlink" title="『算法-ACM 竞赛-图论』二分图-二分图的定义及其判断定"></a>『算法-ACM 竞赛-图论』二分图-二分图的定义及其判断定</h1><h1 id="图论–二分图–二分图的定义及其判断定"><a href="#图论–二分图–二分图的定义及其判断定" class="headerlink" title="图论–二分图–二分图的定义及其判断定"></a>图论–二分图–二分图的定义及其判断定</h1><p><strong>定义：</strong></p>
<p>如果一张无向图的 N 个节点（N&gt;&#x3D;2）可以分成 A B 两个非空子集，其中 A∩B&#x3D;Ø，并且在同一集合内的点之间没有相连的边，则称这张无向图为二分图。A，B 分别成为这个图的左部和右部。</p>
<p><strong>定理：</strong></p>
<p>一张无向图是二分图，当且仅当图中不存在奇环（长度为奇数的环）。</p>
<p><strong>证明：</strong></p>
<p>下面用反证法来证明。<br>假设 X 中的顶点 x1 与 x2 是邻接的，那 UX1，X1X2，X2U 就构成了一个环，这个环的长度为奇数；这与 H 不具有奇环相矛盾。因此，X 中不存在相邻接的顶点。同样可以证明 Y 中也不存在相邻接的顶点。这样，我们就构造出非琐碎组件 H 的两个集合 X 与 Y，X 与 Y 是不相交的，X 中任意两个顶点都不是邻接的；同样 Y 中任意两个顶点也都不是邻接的。因此 H 是二分的。同样可以证明所有其它的 G 的组件都是二分的。因此也就证明了不具有奇环的图是二分图。</p>
<p><strong>匹配：</strong></p>
<p>我们将这种两两不含公共端点的边合集 M 成为成为匹配，而元素最多的边集 M 则称为二分图的最大匹配。当二分图的匹配书等于 2 倍节点数的时候，这个匹配就称为原二分图的完美匹配（完备匹配）</p>
<p><strong>最大匹配：</strong></p>
<p>匈牙利算法（增广路算法）：稍微给你们提一句：</p>
<p><img src="https://img-blog.csdnimg.cn/20191027211957362.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MzYyNzExOA==,size_16,color_FFFFFF,t_70" srcset="/img/loading.gif" lazyload></p>
<pre><code class="hljs">//二分图最大匹配数量
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
#include&lt;iostream&gt;
#include&lt;queue&gt;
#include&lt;vector&gt;
#include&lt;cmath&gt;
#include&lt;algorithm&gt;
using namespace std;
const int N=505;
int line[N][N];
int girl[N],used[N];
int k,m,n;
bool found(int x)
&#123;
    for(int i=1; i&lt;=n; i++)
    &#123;
        if(line[x][i]&amp;&amp;!used[i])
        &#123;
            used[i]=1;
            if(girl[i]==0||found(girl[i]))
            &#123;
                girl[i]=x;
                return 1;
            &#125;
        &#125;
    &#125;
    return 0;
&#125;
int main()
&#123;
    int x,y;
    while(scanf(&quot;%d&quot;,&amp;k)&amp;&amp;k)
    &#123;
        scanf(&quot;%d %d&quot;,&amp;m,&amp;n);
        memset(line,0,sizeof(line));
        memset(girl,0,sizeof(girl));
        for(int i=0; i&lt;k; i++)
        &#123;
            scanf(&quot;%d %d&quot;,&amp;x,&amp;y);
            line[x][y]=1;
        &#125;
        int sum=0;
        for(int i=1; i&lt;=m; i++)
        &#123;
            memset(used,0,sizeof(used));
            if(found(i)) sum++;
        &#125;
        printf(&quot;%d\n&quot;,sum);
    &#125;
    return 0;
&#125;
</code></pre>
<p>** 二分图的最佳完美匹配：**</p>
<p>二分图的最佳完美匹配就是在完备匹配的基础上，每条匹配边都有他的权值，要使权值最大化，最大化权值的完备匹配。</p>
<pre><code class="hljs">#include &lt;iostream&gt;
#include &lt;cstring&gt;
#include &lt;cstdio&gt;

using namespace std;
const int MAXN = 305;
const int INF = 0x3f3f3f3f;

int love[MAXN][MAXN];   // 记录每个妹子和每个男生的好感度
int ex_girl[MAXN];      // 每个妹子的期望值
int ex_boy[MAXN];       // 每个男生的期望值
bool vis_girl[MAXN];    // 记录每一轮匹配匹配过的女生
bool vis_boy[MAXN];     // 记录每一轮匹配匹配过的男生
int match[MAXN];        // 记录每个男生匹配到的妹子 如果没有则为-1
int slack[MAXN];        // 记录每个汉子如果能被妹子倾心最少还需要多少期望值

int N;


bool dfs(int girl)
&#123;
    vis_girl[girl] = true;

    for (int boy = 0; boy &lt; N; ++boy) &#123;

        if (vis_boy[boy]) continue; // 每一轮匹配 每个男生只尝试一次

        int gap = ex_girl[girl] + ex_boy[boy] - love[girl][boy];

        if (gap == 0) &#123;  // 如果符合要求
            vis_boy[boy] = true;
            if (match[boy] == -1 || dfs( match[boy] )) &#123;    // 找到一个没有匹配的男生 或者该男生的妹子可以找到其他人
                match[boy] = girl;
                return true;
            &#125;
        &#125; else &#123;
            slack[boy] = min(slack[boy], gap);  // slack 可以理解为该男生要得到女生的倾心 还需多少期望值 取最小值 备胎的样子【捂脸
        &#125;
    &#125;

    return false;
&#125;

int KM()
&#123;
    memset(match, -1, sizeof match);    // 初始每个男生都没有匹配的女生
    memset(ex_boy, 0, sizeof ex_boy);   // 初始每个男生的期望值为0

    // 每个女生的初始期望值是与她相连的男生最大的好感度
    for (int i = 0; i &lt; N; ++i) &#123;
        ex_girl[i] = love[i][0];
        for (int j = 1; j &lt; N; ++j) &#123;
            ex_girl[i] = max(ex_girl[i], love[i][j]);
        &#125;
    &#125;

    // 尝试为每一个女生解决归宿问题
    for (int i = 0; i &lt; N; ++i) &#123;

        fill(slack, slack + N, INF);    // 因为要取最小值 初始化为无穷大

        while (1) &#123;
            // 为每个女生解决归宿问题的方法是 ：如果找不到就降低期望值，直到找到为止

            // 记录每轮匹配中男生女生是否被尝试匹配过
            memset(vis_girl, false, sizeof vis_girl);
            memset(vis_boy, false, sizeof vis_boy);

            if (dfs(i)) break;  // 找到归宿 退出

            // 如果不能找到 就降低期望值
            // 最小可降低的期望值
            int d = INF;
            for (int j = 0; j &lt; N; ++j)
                if (!vis_boy[j]) d = min(d, slack[j]);

            for (int j = 0; j &lt; N; ++j) &#123;
                // 所有访问过的女生降低期望值
                if (vis_girl[j]) ex_girl[j] -= d;

                // 所有访问过的男生增加期望值
                if (vis_boy[j]) ex_boy[j] += d;
                // 没有访问过的boy 因为girl们的期望值降低，距离得到女生倾心又进了一步！
                else slack[j] -= d;
            &#125;
        &#125;
    &#125;

    // 匹配完成 求出所有配对的好感度的和
    int res = 0;
    for (int i = 0; i &lt; N; ++i)
        res += love[ match[i] ][i];

    return res;
&#125;

int main()
&#123;
    while (~scanf(&quot;%d&quot;, &amp;N)) &#123;

        for (int i = 0; i &lt; N; ++i)
            for (int j = 0; j &lt; N; ++j)
                scanf(&quot;%d&quot;, &amp;love[i][j]);

        printf(&quot;%d\n&quot;, KM());
    &#125;
    return 0;
&#125;
</code></pre>

                
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